Dimitre Novatchev writes (to the Haskell mailing list):
In his excellent book Simon Thompson defines scaling of the elements of a sequence of random numbers from the interval [0, 65535] to an interval [s,t] in the following way (top of page 368):
scaleSequence :: Int -> Int -> [Int] -> [Int] scaleSequence s t = map scale where scale n = n `div` denom + s range = t - s + 1 denom = modulus `div` range where modulus = 65536 However, the following expression:
e4000 = (scaleSequence 1 966 ([65231] ++ [1..965]))!!0 evaluates to 974 -- clearly outside of the specified interval.
The function fails in this case because the range doesn't divide the modulus. In the book (I checked the Miranda version), there's a remark after the function that "[The] range of numbers 0 to modulus-1 is split into range blocks, each of the same length." When the range doesn't divide the modulus, it's not possible to get a uniform distribution if you scale each value individually. I assume this is acceptable, otherwise you'll have to discard or combine values from the original sequence. Oleg answers:
We aim to find a function sc(n) defined over integers 0 <= n <= M so that sc(0) -> s (given integral number) sc(M) -> t (given integral number), t>=s and 0<=a < b ==> sc(a) <= sc(b)
Such a function sc(n) is given by the following expression: sc(n) = (n*(t-s)) `div` M + s
(M = modulus - 1) I think a more natural function would be sc(n) = (n * (t-s+1)) `div` modulus + s This satifies the conditions (assuming range = t-s+1 <= modulus), and distributes the values more evenly, in particular when the range divides the modulus. Cheers, Ronny Wichers Schreur